Show that the sum of an AP whose first term is $ a $, the second term $ b $ and the last term $ c $, is equal to $ \frac{(a+c)(b+c-2 a)}{2(b-a)} $

Given:

An AP whose first term is \( a \), the second term \( b \) and the last term is \( c \).

To do:

We have to show that the sum of an AP whose first term is \( a \), the second term \( b \) and the last term \( c \), is equal to \( \frac{(a+c)(b+c-2 a)}{2(b-a)} \).

Solution:

In the given AP,

First term $=a$

Common difference $=b-a$

Last term $a_{n}=c$

$a_{n}=l=a+(n-1) d$

$c=a+(n-1)(b-a)$

$c-a=(n-1)(b-a)$

$(n-1)=\frac{c-a}{b-a}$

$n=\frac{c-a}{b-a}+1$

$n=\frac{c-a+b-a}{b-a}$

$n=\frac{c+b-2 a}{b-a}$

Sum of the $n$ terms of an AP $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{(b+c-2 a)}{2(b-a)}\left[2 a+\left\{\frac{b+c-2 a}{b-a}-1\right\}(b-a)\right]$

$=\frac{(b+c-2 a)}{2(b-a)}\left[2 a+\frac{c-a}{b-a} (b-a)\right]$

$=\frac{(b+c-2 a)}{2(b-a)}(2 a+c-a)$

$=\frac{(b+c-2 a)}{2(b-a)} (a+c)$

Hence proved.